equation

[karliekay]

Jane has a little lemonade stands with a fixed cost of $10 per day and a variable cost of $0.15 per cup. If lemonade can be sold for $0.50 per cup, find the minimum number of cups Jane must sell to make a profit for the day.

I dont know if the equation is y=.5x+10 or y=.5x-10. Am I looking for when she is making over $10?

 

[Deleted member 4993]

Jane has a little lemonade stands with a fixed cost of $10 per day and a variable cost of $0.15 per cup. If lemonade can be sold for $0.50 per cup, find the minimum number of cups Jane must sell to make a profit for the day.

I dont know if the equation is y=.5x+10 or y=.5x-10. Am I looking for when she is making over $10?

In your equation - what is "y" and what is "x"?

 

[soroban]

Hello, karliekay!

We can communicate if you tell us what \(\displaystyle x\) and \(\displaystyle y\) represent.

Jane has a little lemonade stands with a fixed cost of $10 per day and a variable cost of $0.15 per cup.
If lemonade can be sold for $0.50 per cup, find the minimum number of cups Jane must sell to make a profit for the day.

Let's baby-talk our way through this . . .


\(\displaystyle \text{It costs her }\$10\text{ (1000 cents) per day to run the lemonade stand.}\)

\(\displaystyle \text{She pays }15\rlap{/}c\text{ per cup and sells it for }50\rlap{/}c\quad\hdots\;\text{Her profit is }35\rlap{/}c\text{ per cup.}\)

\(\displaystyle \text{If she sells }x\text{ cups of lemonade, she gets }35x\text{ cents.}\)

\(\displaystyle \text{Her profit for the day is: }\;35x - 1000\text{ cents.}\)


\(\displaystyle \text{If she makes a profit for the day: }\;35x - 1000 \:>\:0\)

\(\displaystyle \text{Solve for }x\!:\;\;35x \:>\:1000 \quad\Rightarrow\quad x \:>\:\frac{1000}{35} \:=\:28\tfrac{4}{7}\)


\(\displaystyle \text{Therefore, she must sell at least 29 cups of lemonade.}\)